Markov's inequality states that for any non-negative random variable X and any a > 0, the probability that X is at least a × its mean is at most 1/a. In this con, let p be the proportion of incomes that are over $100,000. Markov's inequality implies:

a) P(X > 100,000) ≤ (Mean of X) / 100,000

b) P(X > 100,000) ≤ (Mean of X) 100,000

c) P(X > 100,000) ≤ 100,000 / (Mean of X)

d) P(X > 100,000) ≤ 100,000 (Mean of X)